bài 3:a)O=AC x BD (x là giao nhá)=> SO \(\perp\) (ABCD)
=> OC=\(a\sqrt{2}\)\(\Rightarrow\widehat{SCO}=60^o\Rightarrow SO=OC.tan60^o=\frac{a\sqrt{6}}{2}\Rightarrow V_{k.chóp}=\frac{1}{3}SO.S_{ABCD}=\frac{1}{3}.a\frac{\sqrt{6}}{2}.a^2=\frac{a^3\sqrt{6}}{6}\)
b) \(\Delta SAC\)có \(\widehat{SCA=60^o}\)=> \(\Delta SAC\)đều
AE\(\perp\)SC=> AE=\(\frac{a\sqrt{6}}{2}\)
AExSO=G => G là trọng tâm \(\Delta SAC\)=> \(\frac{SG}{SO}\)=\(\frac{2}{3}\)
\(\hept{\begin{cases}BD\perp SO\\BD\perp AC\end{cases}\Rightarrow BD\perp\left(SAC\right)\Rightarrow BD\perp SC}\)
(AMEN)\(\perp\)SC => MN \(\perp\)SC => MN //BD => \(\frac{MN}{BD}=\frac{SG}{SO}=\frac{2}{3}\Rightarrow MN=\frac{2}{3}BD=\frac{2a\sqrt{2}}{3}\)
\(S_{AMEN}=\frac{1}{2}MN.AE=\frac{1}{2}.\frac{2a\sqrt{2}}{3}.\frac{a\sqrt{6}}{2}=\frac{a^2\sqrt{3}}{3}\)
\(\frac{V_{SAMEN}}{V_{SABCD}}=\frac{SM}{SB}.\frac{SE}{SC}.\frac{SN}{SD}=\frac{2}{3}.\frac{1}{2}.\frac{2}{3}=\frac{2}{9}\)
\(\Rightarrow V_{SAMEN}=\frac{2}{9}.\frac{a^3\sqrt{6}}{6}=\frac{a^3\sqrt{6}}{27}\)
phần trả lời bên dưới là câu 4
I*AB=> SI\(\perp\)AB
SI=\(SI=\frac{AB\sqrt{3}}{2}=\frac{a\sqrt{3}}{2}\)
\(V_{k.chop}=\frac{1}{3}.\frac{a\sqrt{3}}{2}.a^2=\frac{a^3\sqrt{3}}{4}\)
b) Kẻ IK//DM(K\(\in\)AD)
Kẻ KH\(\perp\)DM(H\(\in\)DM)
=> d(I,DM)=d(K,DM0=KH
\(\Delta IAK~\Delta DCM\Rightarrow AK=\frac{1}{2}CM=\frac{a}{6}\)=> KD=5a/6
\(cos\widehat{ADM}=cos\widehat{DMC}=\frac{CM}{DM}=\frac{\frac{a}{3}}{\frac{a\sqrt{10}}{3}}=\frac{1}{\sqrt{10}}\)
=> KH=KDsin\(\widehat{ADM}\)=\(\sqrt{1-\cos\widehat{ADM}^2}=\frac{5a}{6}.\frac{3}{\sqrt{10}}=\frac{a\sqrt{10}}{4}\)
d(S,DM)=\(\sqrt{SI^2+d\left(I,DM\right)^2}=\frac{a\sqrt{22}}{4}\)
